STRUCTURE OF Eu-151 AND Eu-153
By Prof Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. STRUCTURES OF EUROPIUM ISOTOPES Naturally occurring europium (Eu) is composed of 2 isotopes, Eu-151 and Eu-153, with Eu-153 being the most abundant (52.2% natural abundance). While Eu-153 is stable, Eu-151 was recently found to be unstable and to undergo alpha decay with half-life of (4.62 ± 0.95(stat.) ± 0.68(syst.)) × 1018 y. Comparing the structures of europium of 63 protons (odd number) with those of Samarium of 62 protons (even number) we see that the nuclides of Europium break the high symmetry of Samarium . (See my STRUCTURE OF Sm-144...Sm-154 ). After a careful analysis of this comparison I discovered that the additional vertical p63n63 forms with the n62p62 the alpha particle on the right side which gives 2(n) of opposite spins. Then, for symmetrical arrangements the n39p39 changes the spin from S =-1 to S =0 giving S = +1. Particularly it is moved from the square of negative spins to the p61n61 with S =0 for making the symmetrical alpha particle (on the left side) which gives also 2(n) of opposite spins. So under these arrangements the number N of blank positions is given by The p37n37 gives 3n of strong bonds with negative spins. The square gives 4n with positive spins. The first and the sixth plane give 4(n) of weak bonds with opposite spins. The second and the fifth plane give 4{n} with three bonds per neutron and 8n. The third and the fourth plane give 4(n) of weak bonds with opposite spins Also the two symmetrical alpha particles give at the third and the fourth plane 4(n) of opposite spins That is N= 4{n} +15n + 12(n) = 31 blank positions able to receive 16 extra neutrons of positive spins and 15 extra neutrons of negative spins. ' ' STRUCTURE OF Eu-151 AND Eu-153 WITH S = +5/2 Since the Eu-151 of 25 extra neutrons has S =+5/2 we conclude that it has 14 extra neutrons of positive spins and 11 extra neutrons of negative spins. That is S = +1 + 14(+1/2) + 11(-1/2) = +5/2 Whereas the Eu-153 with S = +5/2 of 27 extra neutrons has 15 extra neutrons of positive spins and 12 extra neutrons of negative spins. That is S = +1 +15(+1/2) + 12(-1/2) = +5/2 DIAGRAM OF EUROPIUM-126 FORMING 31 BLANK POSITIONS In this structure you can see the six horizontal planes of opposite spins like the +HP1, -HP2, +Hp3, -HP4 +HP5 and -HP6. You can see also the square of negative spins ( -HSQ) with the one p37n37 because the p39n39 as a vertical system with s =0 is absent. Particularly it went to p61n61 for making a symmetrical alpha particele. Note that this change of spin gives a total S = +1. Whereas at the square of positive spins (+HSQ) you can see both p38n38 and p40n40. Here the additional p63 and n63 are shown near the n62 and p62 respectively. So for symmetrical arrangements the n39 and p39 are shown near the p61 and n61 respectively. Note that the p47n47 along with the p48n48 make inside the two symmetrical alpha particles of opposite spins . But you cannot see the p49n49, the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' n40......p40........n' ' n......... p38.......n38 +HSQ' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 -HP6' ' n........p29.........n10.........p10…… n30 ' ' n29…...p9..........n9 …….p30.........n +HP5' ' n61....p47.......n27.........p8..........n8.........p28........... n48......p62' ' p39....n45...........p27........n7.........p7........n28..........p46...........n63 -HP4 ' ' p61......n47.........p25.........n6.........p6..........n26...........p48.....n62' ' n39....p45..........n25…..p5..........n5……….p26.......n46 .........p63 +HP3' ' n23………p4..........n4………….p24..............n' ' n......p23…….....n3………..p3………..n24 -HP2' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22] +Hp1' ' n.........p37......n37 ' ' -HSQ' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' Here you see the 2(n) of weak horizontal bonds ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE' Here we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4.........n4.........p24..........n ' ' n.......p23........n3........p3.........n24' ' {n}...... p13......n13 ' ' n' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' Here the first 2(n) of horizontal bonds fill the blank positions of the central parallelepiped, while the second 2(n) of horizontal bonds are formed by the additional alpha particles. Using this top view of the third plane you can see the following characteristics of the fundamental shapes formed by the nucleons of the central parallelepiped as The p5n5 and n6p6 create the small horizontal square of Mg-24 for creating the central parallelepiped of the alpha particle nuclei. The n15p15 and p16n16 create the first small horizontal rectangle. The p25n25 and p26n26 create the second small horizontal rectangle. The p41, n42, n43 and p44 make the great horizontal square of the great central parallelepiped. The p45, n46, n47 and p48 form the first great horizontal rectangle. The p49, n50, p51 and n52 form the second great horizontal rectangle. ' ' ' (n)........p58....... n50.......p51....n60 ' ' (n) p53........n42........p16......n16......p44.........n54 ' p61 n47........p25........n6........p6........n26.........p48 n62' ' n39 p45........n25........p5........n5........p26........ n46 p63' ' n55........p41.......n15.......p15.......n43...... .p56 (n)' ' n57.......p49.......n52...... p59........(n)' TOP WIEW OF THE UP HORIZONTAL SQUARE Here the 2n near p38 have strong bonds with p31 and p35 respectively. Whereas the 2n near p40 have the strong bonds with p32 and p36 respectively. ' n' ' n40......p40.......n ' n.....p38.......n38 ' n' Category:Fundamental physics concepts